Bài 1 Tính1)\(\sqrt{\left(-0,3\right)^2}\)2)\(-\frac{1}{2}\sqrt{\left(0,3\right)^2}\)3)\(\sqrt{a^{10}}\)(a>0)4)\(\sqrt{\left(2-x^2\right)}\)(x<2)5)\(\sqrt{x^2+2x+1}\)6)\(\sqrt{\left(1-\sqrt{2}\right)^2}\)7)\(\sqrt{11+6\sqrt{2}}\)8)\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}\)9)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)Làm hộ e, mk vs ạ e, mk xin cảm ơn bn, a, cj...
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Bài 1 Tính
1)\(\sqrt{\left(-0,3\right)^2}\)
2)\(-\frac{1}{2}\sqrt{\left(0,3\right)^2}\)
3)\(\sqrt{a^{10}}\)(a>0)
4)\(\sqrt{\left(2-x^2\right)}\)(x<2)
5)\(\sqrt{x^2+2x+1}\)
6)\(\sqrt{\left(1-\sqrt{2}\right)^2}\)
7)\(\sqrt{11+6\sqrt{2}}\)
8)\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}\)
9)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
Làm hộ e, mk vs ạ e, mk xin cảm ơn bn, a, cj ạ
Câu 1 =3/10
\(1,\sqrt{\left(-0,3\right)^2}=\sqrt{0,09}=0,3\)
\(2,-\frac{1}{2}\sqrt{\left(0,3\right)^2}=-\frac{1}{2}.0,3=-0,15\)
\(3,\sqrt{a^{10}}=\sqrt{\left(a^5\right)^2}=a^5\left(a\ge0\right)\)
\(4,\sqrt{\left(2-x\right)^2}=\left|2-x\right|=2-x\left(x\le2\right)\)
\(5,\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)
\(6,\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)(Vì \(1< \sqrt{2}\))
\(7,\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
\(8,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)
\(=-2\)
\(9,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)